You are here:

Physics/Kepler's Laws

Advertisement


Question
Problem
Problem  
(Given: The value of the universal gravitational constant is 6.672e−11 N m2/kg2)

Objects with masses of 224 kg and 614 kg are separated by 0.369 m. A 61.1 kg mass is placed midway between them.

a) Find the magnitude of the net gravitational force exerted by the two larger masses on the 61.1 kg mass. Answer in units of N

b) Leaving the distance between the 224 kg and the 614 kg masses fixed, at what distance from the 614 kg mass (other than infinitely remote ones) does the 61.1 kg mass experience a net
force of zero? Answer in units of m

Answer
The gravitational force between two bodies is given by:
Fg=G*m1*m2/R^2
In this case the gravitational force of the left mass m1 on the center mass m3 will be:
Fg13=G*m1*m3/R13^2=(6.67x10^-11)*224*61.1/(.369/2)^2=2.68x10^-5N left
While the gravitational force on the center mass by the right mass will be:
Fg23=G*m2*m3/R23^2=2=(6.67x10^-11)*614*61.1/(.369/2)^2=7.35x10^-5 right
Since force is a vector subtract to determine the net force,on th center mass
Fgnet=Fg23-Fg13=4.67x10^-5N right
For the center mass to feel zero net force these two gravitational forces must be equal:
Fg13=Fg23
G*m1*m3/R13^2=G*m2*m3/R3023^2
Simplifying:
m1/R13^2=m2/R23^2
Solve for R23 in terms of R13:
m2/m1=R23^2/R13^2
R23=R13*sqrt(m2/m1)=R13*sqrt(614/224)=1.66*R13
Since these two distance must add up to 0.369m:
R13+R23=R13+1.66*R13=2.66*R13=0.369
Therefore, R13 becomes:
R13=0.369/1.66=0.139m
And:
R23=0.369-R13=0.369-0.139=0.230m

Physics

All Answers


Answers by Expert:


Ask Experts

Volunteer


James J. Kovalcin

Expertise

I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

©2016 About.com. All rights reserved.